# proof of equivalent definitions of analytic sets for Polish spaces

Let $A$ be a nonempty subset of a Polish space  $X$. Then, letting $\mathcal{N}$ denote Baire space  and $Y$ be any uncountable Polish space, we show that the following are equivalent      .

1. 1.

$A$ is $\mathcal{F}$-analytic  (http://planetmath.org/AnalyticSet2).

2. 2.

$A$ is the projection (http://planetmath.org/GeneralizedCartesianProduct) of a closed subset of $X\times\mathcal{N}$ onto $X$.

3. 3.

$A$ is the image (http://planetmath.org/DirectImage) of a continuous function   $f\colon Z\to X$ for some Polish space $Z$.

4. 4.

$A$ is the image of a continuous function $f\colon\mathcal{N}\to X$.

5. 5.

$A$ is the image of a Borel measurable function $f\colon Y\to X$.

6. 6.

$A$ is the projection of a Borel subset of $X\times Y$ onto $X$.

(1) implies (2): Let $\mathcal{F}$ be the paving consisting of closed subsets of $X$. The collection  of $\mathcal{K}$-analytic sets  contains the Borel $\sigma$-algebra of $X$ (see countable unions and intersections of analytic sets are analytic) and, as the analytic sets are given by a closure operator   (http://planetmath.org/AnalyticSetsDefineAClosureOperator) it follows that it contains all analytic subsets of $X$. So, any analytic subset $A$ of $X$ is $\mathcal{F}$-analytic. Then, there is a closed $S\subseteq\mathcal{N}$ and a function $\theta\colon\mathbb{N}\to\mathcal{F}$ such that

 $A=\bigcup_{s\in S}\bigcap_{n=1}^{\infty}\theta(s_{n})$

(see proof of equivalent definitions of analytic sets for paved spaces). For each $m,n\in\mathbb{N}$ let $K_{m,n}$ denote the closed subset of $s\in\mathcal{N}$ with $s_{n}=m$. Then, we can rearrange the above expression to get $A=\pi_{X}(B)$ where $\pi_{X}\colon X\times\mathcal{N}\to X$ is the projection map and

 $B=(X\times S)\cap\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}\theta(m)\times K% _{m,n}.$

It is easily seen that $\bigcup_{m}\theta(m)\times K_{m,n}$ is a closed subset of $X\times\mathcal{N}$ for each $n$, and therefore $B$ is closed, as required.

(2) implies (3): Suppose that $A=\pi_{X}(S)$ for a closed subset $S$ of $X\times\mathcal{N}$, where $\pi_{X}\colon X\times\mathcal{N}\to X$ is the projection map. As the product   of Polish spaces is Polish, and every closed subset of a Polish space is Polish, then $S$ will be a Polish space under the subspace topology. So, we can take $Z=S$ and let $f\colon Z\to X$ be the restriction  of $\pi_{X}$ to $Z$.

(3) implies (4): Suppose that $A$ is the image of a continuous function $g\colon Z\to X$, for a Polish space $Z$. As Baire space is universal for Polish spaces, there exists a continuous and onto () function $h\colon\mathcal{N}\to Z$. The result follows by taking $f=g\circ h$.

(4) implies (5): Suppose that $A$ is the image of a continuous function $g\colon\mathcal{N}\to X$. Since uncountable Polish spaces are all Borel isomorphic (see Polish spaces up to Borel isomorphism), there is a Borel isomorphism $h\colon Y\to\mathcal{N}$. The result follows by taking $f=g\circ h$.

(5) implies (6): Suppose that $A$ is the image of a Borel measurable function $f\colon Y\to X$, and let $\Gamma$ be its graph (http://planetmath.org/Graph2)

 $\Gamma\equiv\left\{(f(y),y)\colon y\in Y\right\}\subseteq X\times Y.$

The projection of $\Gamma$ onto $X$ is equal to $f(Y)=A$, so the result will follow once it is shown that $\Gamma$ is a Borel set.

Choose a countable  and dense subset $\{x_{1},x_{2},\ldots\}$ of $X$, and let $d$ be a metric generating the topology  on $X$. Then, for integers $m,n\geq 1$, denote the open ball  about $x_{m}$ of radius $1/n$ by $B_{m,n}$. Since the $x_{m}$ form a dense set, $\bigcup_{m}B_{m,n}=X$ for each $n$. Let us define

 $\Gamma_{n}\equiv\bigcup_{m=1}^{\infty}B_{m,n}\times f^{-1}(B_{m,n})\subseteq X% \times Y,$

which contains $\Gamma$. Furthermore, since $f^{-1}(B_{m,n})$ are Borel, $\Gamma_{n}$ are Borel sets. Suppose that $(x,y)\in\bigcap_{n}\Gamma_{n}$. Then, for each $n$, there is an $m$ such that $x\in B_{m,n}$ and $y\in f^{-1}(B_{m,n})$. So,

 $d(x,f(y))\leq d(x,x_{m})+d(x_{m},f(y))\leq 2/n.$

This holds for all $n$, showing that $y=f(x)$ and so $(x,y)\in\Gamma$. We have shown that $\Gamma=\bigcap_{n}\Gamma_{n}$ is Borel, as required.

(6) implies (1): This is an immediate consequence of the result that projections of analytic sets are analytic.

Title proof of equivalent definitions of analytic sets for Polish spaces ProofOfEquivalentDefinitionsOfAnalyticSetsForPolishSpaces 2013-03-22 18:48:48 2013-03-22 18:48:48 gel (22282) gel (22282) 5 gel (22282) Proof msc 28A05